(I'm surprised this question or a very similar one hasn't been asked, so I wouldn't be surprised if I just failed to find a question with my search that does answer this question.)

I dissolved some $\ce{K3PO4}$ in water a while back, and the solution became very hot. I commented on this to my labmate, and he said "So that means it's more soluble at low temperature, right?" I evaded response because I disagreed but didn't have a thought-out response yet.

Phosphoric acid's third pKa is pretty high, so the second reaction is pretty significant. This, I would say, is why this particular reaction is so exothermic. So, in this case, I don't really think the heat plays into the solubility of potassium phosphate.

Sometimes, though, solvation comes with a positive change in temperature. For instance, the enthalpy of solvation of sodium iodide is $\mathrm{-7.53\ kJ/mol}$. Sodium iodide, though, is still more soluble at high temperatures. I could be missing something here, though.

In the cases where it is specifically the solvation reaction that causes a temperature change, is it the true that lower temperatures always increase solubility of salts with exothermic solvation reactions (and vice-versa)?

I'd say that the answer is no because the solvation should be dependent on $\Delta H - T\Delta S$ and, since the enthalpy change half of that formula is temperature independent, as $T$ goes up, solvation should be favored by any solvation that has a positive change in entropy. However, I can't really think of any solvation that has a negative change in entropy.

$\begingroup$I don't think it matters what goes on with the base forming an acid or all that. What matters is that the overall process--whatever happens--is exothermic. Thus, if you increase the temperature, you are favoring the side that "contains" more heat, the "reactants." Thus, the equilibrium constant with an exothermic forward reaction decreases as temperature increases, and in this case, solubility decreases; it is more soluble at low temperatures.$\endgroup$
– Yunfei MaApr 12 '16 at 0:38

$\begingroup$@YunfeiMa Well then why is potassium phosphate more soluble at high temperatures if it gets too hot for me to touch as it dissolves?$\endgroup$
– SendersReagentApr 12 '16 at 0:55

$\begingroup$CO2 has a negative solubility curve in water (Lower solubility with higher temperature). (And an exothermic heat of reaction). But feel free to have a pop anyways.$\endgroup$
– LighthartApr 12 '16 at 4:08

2 Answers
2

Now it is pretty clear that exothermic reaction means negative $\Delta H$, which in turn means positive coefficient in $-{\Delta H\over RT}$, which consequently is a decreasing function, as is the whole exponential expression. The entropy change (or indeed the sign of it) does not matter.

The example of $\ce{CaCl2}$ is irrelevant, for anhydrous $\ce{CaCl2}$ is never the phase in equilibrium with the solution. Roughly speaking, one simply cannot just up and dissolve $\ce{CaCl2}$. In effect, there would be two reactions: first the hydration, which produces $\ce{CaCl2.6H2O}$ (that's the exothermic part), and then the dissolution of that.

The implicit approximation we relied upon is that our $\Delta H$ and $\Delta S$ won't change much throughout the dissolution process. This fails when it comes to insanely well soluble things (think of $\ce{NaOH},\;\ce{KOH}$) where by the time we reach equilibrium, we might even have more solute than water. Naturally, the properties of such solution would be quite different from those of pure water.

$\begingroup$So, based on this, the case of $\ce{K3PO4}$ must be exothermic due to the acid-base reaction or something else other than the actual solvation process because its solubility goes up with temperature, right?$\endgroup$
– SendersReagentApr 12 '16 at 8:13

1

$\begingroup$I'm not sure. This might fall in the last category (where the assumption of constant $\Delta H$ fails).$\endgroup$
– Ivan NeretinApr 12 '16 at 8:23

$\begingroup$Good point. It is very soluble.$\endgroup$
– SendersReagentApr 12 '16 at 8:23

$\begingroup$Unless of course you were purposely curt, in which case: kthxbi. I get the feeling, however, that you just didn't read the question fully.$\endgroup$
– SendersReagentApr 12 '16 at 4:08

$\begingroup$Huh? You asked "... is it the true that lower temperatures always increase solubility of salts with exothermic solvation reactions?... " I gave a counter example which disproves the premise. So the answer is no.$\endgroup$
– MaxWApr 12 '16 at 4:11

$\begingroup$And used exactly the reaction I mentioned where I said that I don't think it's the solvation that causes the exotherm in that case, but rather the acid-base reaction that occurs afterwards. So the answer is unaffected by your response. Unless you care to describe how it isn't due to the acid-base reaction in your answer.$\endgroup$
– SendersReagentApr 12 '16 at 4:15

$\begingroup$Ok, you're right I didn't read and process all you question. I found a different black swan - $\ce{CaCL2}$. So it can't be "always."$\endgroup$
– MaxWApr 12 '16 at 5:12